If the matrix $A = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$, then the value of det $(A^2 - 2A)$ is equal to

Given $A = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$

Finding $A^2$:

$A^2 = A \times A$

$= \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$

$= \begin{bmatrix} (1)(1) + (3)(2) & (1)(3) + (3)(1) \\ (2)(1) + (1)(2) & (2)(3) + (1)(1) \end{bmatrix}$

$= \begin{bmatrix} 7 & 6 \\ 4 & 7 \end{bmatrix}$

Finding $2A$:

$2A = 2 \times \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$

$= \begin{bmatrix} 2 & 6 \\ 4 & 2 \end{bmatrix}$

Finding $A^2 - 2A$:

$A^2 - 2A = \begin{bmatrix} 7 & 6 \\ 4 & 7 \end{bmatrix} - \begin{bmatrix} 2 & 6 \\ 4 & 2 \end{bmatrix}$

$= \begin{bmatrix} 7-2 & 6-6 \\ 4-4 & 7-2 \end{bmatrix}$

$= \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}$

Finding the determinant:

$\text{det}(A^2 - 2A) = \begin{vmatrix} 5 & 0 \\ 0 & 5 \end{vmatrix}$

$= (5)(5) - (0)(0)$

$= 25$

Therefore, $\text{det}(A^2 - 2A) = 25$